Question

the hubble space telescope was deployed on April 24, 1990, by the space shuttle discovery. A model for the velocity of the shuttle during this mission, from liftoff at t=0 until the solid rocket boosters were jettisoned at t=126 is given by v(t)=0.001302t^3 - 0.09029t^2 +23.61t - 3.083 (in feet per second). Using this model, estimate the absolute maximum and minimum values of the acceleration of the shuttle between liftoff and the jettisoning of the boosters.

Answer 1

hey kayla !
good to see you here again.

so the acceleration is the derivative with respect to time of the velocity :

a(t) = 0.003906t^2-0.18058t+23.61 (this is in feet per second squared)

in order to find maximum or minimum of a function we need to take its derivative and equate to zero.

a'(t) = 0
a'(t) = 0.007812t-0.18058
so t = 23.12 sec
we need to find a(23.12).
and at this time the acceleration was minimal because you see a(t) is a parabola with a minimum.

since a(t) is a parabola with a minimum, the maximum value of the acceleration will be in one of the two end points ( t = 0 or t=126)
can you tell which ?